Part 3: Converting Type I and Type II integrals   

We can convert a type I or type II integral into different coordinates by first converting into a double integral and then using (1).      

EXAMPLE 3    Use the transformation T( u,v) = á u,u+v ñ to evaluate the iterated integral

ó õ 2 1  ó õ x+3 x+2   dy dx xy-x2

Solution: To do so, we notice that as a double integral we have

ó õ 2 1  ó õ x+3 x+2   dy dx xy-x2  =  ó õ ó õ R   1 xy-x2   dA

where R is the region bounded by x = 1, x = 2, y = x+2 and y = x+3. Since T( u,v) = á u,u+v ñ implies that x = u, y = u+v, we have

y = x+2    Þ     u+v = u+2    Þ     v = 2 y = x+3    Þ     u+v = u+3    Þ     v = 3 x = 1    Þ     u = 1 x = 2    Þ     u = 2

Thus, the pullback of R is u = 1, u = 2, v = 2, v = 3.

            To compute the Jacobian, we notice that x = u, y = u+v implies that

¶( x,y) ¶( u,v) =  ¶x ¶u  ¶y ¶v -  ¶x ¶v  ¶y ¶u = 1·1-1·0 = 1

Thus, dA = 1dudv, so that we transform the double integral into

ó õ ó õ R   1 xy-x2 dA = ó õ 2 1  ó õ 3 2   1 u( u+v) -u2   du dv = ó õ 2 1  ó õ 3 2   1 u2+uv-u2   du dv = ó õ 2 1  ó õ 3 2   1 Ö uv  du dv

As a result, we have

ó õ 2 1  ó õ x+3 x+2   dy dx xy-x2 = ó õ 2 1  ó õ 3 2  u-1/2v-1/2dudv = 4Ö2Ö3-8-4Ö3+4Ö2

               

EXAMPLE 4    Use T( u,v) = áu²,v ñ to evaluate

ó õ 1 0  ó õ Öx 0  yeÖx dydx

Solution: The region R of integration is bounded the curves x = 0,  x = 1, y = 0, and y = Öx. Since x = u² and y = v, we have

x = 0    Þ     u = 0 x = 1    Þ     u = 1 y = 0    Þ     v = 0 y = Öx    Þ     v = u

Moreover, x = u² and y = v implies that

¶( x,y) ¶( u,v) = æ è  ¶ ¶u u2 ö ø æ è  ¶  ¶v v ö ø - æ è  ¶  ¶v u2 ö ø æ è  ¶ ¶u v ö ø = 2u

Thus, dA = 2ududv, so that

ó õ 1 0  ó õ Öx 0  yeÖx dydx = ó õ ó õ R  yeÖx  dA = ó õ 1 0  ó õ u 0  veu  2u  dvdu = ó õ 1 0  u3eudu = -2e+6